18.4 Support Functions
Definition 18.4.1 (Support Function).label Let $\dpn{E, F}{\lambda}$ be a duality over $\real$ and $A \subset E$ be non-empty, then the mapping
the support function of $A$ with respect to $\dpn{E, F}{\lambda}$.
Definition 18.4.2 (Indicator Function).label Let $E$ be a vector space over $\real$ and $A \subset E$, then the mapping
is the infinity characteristic function/indicator function of $A$.
Lemma 18.4.3.label Let $\dpn{E, F}{\lambda}$ be a duality over $\real$, then:
- (1)
For any non-empty $A \subset E$, let $\ol{\conv}(A)$ be the $\sigma(E, F)$-closed convex hull of $A$, then $H_{A} = H_{\ol{\conv}(A)}$.
- (2)
For any non-empty $A, B \subset E$, $H_{A} \le H_{B}$ if and only if $A$ is contained in the $\sigma(E, F)$-closed convex hull of $B$.
- (3)
For any non-empty $A \subset E$, $I_{A}^{*} = H_{A}$.
- (4)
For any $A \subset E$ non-empty, $\sigma(E, F)$-closed, and convex, $H_{A}^{*} = I_{A}$.
Proof. (1): Since $\ol{\conv}(A) \supset A$, $H_{A} \le H_{\ol{\conv}(A)}$. On the other hand, for each $\phi \in F$, $A \subset \bracs{\phi \le H_A(\phi)}$. Since $\bracs{\phi \le H_A(\phi)}$ is a $\sigma(E, F)$-closed convex set, it contains $\ol{\conv}(A)$. Thus $\ol{\conv}(A) \subset \bracs{\phi \le H_A(\phi)}$ and $H_{\ol{\conv}(A)}(\phi) \le H_{A}(\phi)$.
(2): Using (1), assume without loss of generality that $B$ is $\sigma(E, F)$-closed and convex. Suppose that $H_{A} \le H_{B}$, then $A \subset \bigcap_{\phi \in F}\bracs{\phi \le H_B(\phi)}$. By the Hahn-Banach Theorem, $B = \bigcap_{\phi \in F}\bracs{\phi \le H_B(\phi)}$, so $A \subset B$.
(3): Let $\phi \in F$, then since $I_{A}|_{A^c}= \infty$,
(4): Given that $A$ is $\sigma(E, F)$-closed and convex, $I_{S}$ is convex and lower semicontinuous. By the Fenchel-Moreau Theorem and (3), $I_{A} = I_{A}^{**}= H_{A}^{*}$.$\square$
Theorem 18.4.4.label Let $\dpn{E, F}{\lambda}$ be a duality over $\real$, $f: X \to (-\infty, \infty]$ be a $\sigma(E, F)$-lower semicontinuous, subadditive, and positively homogeneous function with $f(0) = 0$, and
then:
- (1)
$f^{*} = I_{\Sigma}$.
- (2)
$\Sigma$ is the unique non-empty $\sigma(F, E)$-closed convex subset of $F$ such that $f = H_{\Sigma}$.
- (3)
$\Sigma$ is equicontinuous if and only if there exists $U \in \cn_{E}(0)$ such that $\sup_{x \in U}f(x) < \infty$.
Conversely,
- (3)
For any non-empty $\Sigma \subset A$, $H_{\Sigma}$ is a lower semicontinuous, subadditive, and positively homogeneous function with $H_{\Sigma}(0) = 0$.
Proof, [Theorem 4.25, Cla13]. (1): Let $\phi \in \Sigma$, then since $\dpn{x, \phi}{\lambda}\le f(x)$ for all $x \in E$,
where the supremum is achieved at $0$. On the other hand, if $\phi \in F \setminus \Sigma$, then there exists $x \in E$ such that $\dpn{x, \phi}{\lambda}- f(x) > 0$. In which case, by positive homogeneity of $f$,
Thus $f^{*}= I_{\Sigma}$.
(2): By Lemma 18.3.7, $I_{\Sigma} \ne \infty$, so $\Sigma \ne \emptyset$. Since
is an intersection of $\sigma(F, E)$-closed and convex sets, it is also $\sigma(F, E)$-closed.
Now, let $x, y \in E$ and $t \in [0, 1]$, then since $f$ is subadditive and positively homogeneous,
so $f$ is convex. Given that $f$ is also $\sigma(E, F)$-lower semicontinuous, the Fenchel-Moreau Theorem and (4) of Lemma 18.4.3 imply that
By (2) of Lemma 18.4.3, $\Sigma$ is the unique closed $\sigma(F, E)$-convex set such that $f = H_{\Sigma}$.
(3): Let $U \in \cn_{E}(0)$ be circled such that $M = \sup_{x \in U}f(x) < \infty$, then
Since $U$ is circled, $\bigcup_{y \in \Sigma}\dpn{M^{-1}U, y}{\lambda}\subset \ol{B_\real(0, 1)}$, so $\Sigma$ is equicontinuous by Proposition 11.14.1.
Conversely, if $\Sigma$ is equicontinuous, then there exists $U \in \cn_{E}(0)$ such that $M = \sup_{y \in \Sigma, x \in U}\dpn{x, y}{\lambda}< \infty$. In which case, (2) implies that
(4): By Proposition 5.22.3, $H_{\Sigma}$ is lower semicontinuous.
Let $x, y \in E$ and $\mu > 0$, then
and
so $H_{\Sigma}$ is subadditive and positively homogeneous.$\square$
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